1.4 problem Example 6

Internal problem ID [2297]
Book : Differential equations and their applications, 3rd ed., M. Braun
Section : Section 1.2. Page 6
Problem number : Example 6
Date solved : Thursday, October 17, 2024 at 02:22:55 AM
CAS classification : [_separable]

Solve

\begin{align*} y^{\prime }+2 t y&=t \end{align*}

With initial conditions

\begin{align*} y \left (1\right )&=2 \end{align*}

1.4.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as

\begin{align*} y^{\prime } + q(t)y &= p(t) \end{align*}

Where here

\begin{align*} q(t) &=2 t\\ p(t) &=t \end{align*}

Hence the ode is

\begin{align*} y^{\prime }+2 t y = t \end{align*}

The domain of \(q(t)=2 t\) is

\[ \{-\infty <t <\infty \} \]

And the point \(t_0 = 1\) is inside this domain. The domain of \(p(t)=t\) is

\[ \{-\infty <t <\infty \} \]

And the point \(t_0 = 1\) is also inside this domain. Hence solution exists and is unique.

1.4.2 Solved as first order linear ode

Time used: 0.140 (sec)

In canonical form a linear first order is

\begin{align*} y^{\prime } + q(t)y &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=2 t\\ p(t) &=t \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int 2 t d t}\\ &= {\mathrm e}^{t^{2}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu y\right ) &= \left (\mu \right ) \left (t\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (y \,{\mathrm e}^{t^{2}}\right ) &= \left ({\mathrm e}^{t^{2}}\right ) \left (t\right ) \\ \mathrm {d} \left (y \,{\mathrm e}^{t^{2}}\right ) &= \left (t \,{\mathrm e}^{t^{2}}\right )\, \mathrm {d} t \\ \end{align*}

Integrating gives

\begin{align*} y \,{\mathrm e}^{t^{2}}&= \int {t \,{\mathrm e}^{t^{2}} \,dt} \\ &=\frac {{\mathrm e}^{t^{2}}}{2} + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{t^{2}}\) gives the final solution

\[ y = \frac {1}{2}+c_1 \,{\mathrm e}^{-t^{2}} \]

Solving for the constant of integration from initial conditions, the solution becomes

\begin{align*} y = \frac {1}{2}+\frac {3 \,{\mathrm e} \,{\mathrm e}^{-t^{2}}}{2} \end{align*}

(a) Solution plot
\(y = \frac {1}{2}+\frac {3 \,{\mathrm e} \,{\mathrm e}^{-t^{2}}}{2}\)

(b) Slope field plot
\(y^{\prime }+2 t y = t\)

1.4.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y \left (t \right )+2 t y \left (t \right )=t , y \left (1\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=-2 t y \left (t \right )+t \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d t}y \left (t \right )}{2 y \left (t \right )-1}=-t \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {\frac {d}{d t}y \left (t \right )}{2 y \left (t \right )-1}d t =\int -t d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (2 y \left (t \right )-1\right )}{2}=-\frac {t^{2}}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )=\frac {{\mathrm e}^{-t^{2}+2 \mathit {C1}}}{2}+\frac {1}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=2 \\ {} & {} & 2=\frac {{\mathrm e}^{2 \mathit {C1} -1}}{2}+\frac {1}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\frac {1}{2}+\frac {\ln \left (3\right )}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =\frac {1}{2}+\frac {\ln \left (3\right )}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {3 \,{\mathrm e}^{-\left (t -1\right ) \left (t +1\right )}}{2}+\frac {1}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {3 \,{\mathrm e}^{-\left (t -1\right ) \left (t +1\right )}}{2}+\frac {1}{2} \end {array} \]

1.4.4 Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

1.4.5 Maple dsolve solution

Solving time : 0.012 (sec)
Leaf size : 17

dsolve([diff(y(t),t)+2*t*y(t) = t, 
       op([y(1) = 2])],y(t),singsol=all)
 

\[ y = \frac {1}{2}+\frac {3 \,{\mathrm e}^{-\left (t -1\right ) \left (t +1\right )}}{2} \]

1.4.6 Mathematica DSolve solution

Solving time : 0.037 (sec)
Leaf size : 22

DSolve[{D[y[t],t]+2*t*y[t]==t,y[1]==2}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to \frac {3 e^{1-t^2}}{2}+\frac {1}{2} \]